What the heck my teacher did in the solution!

I got the problem wrong, and I am trying to understand what my teacher did. Particularly, on the right side of his solution, he took the θ acceleration to be zero, because there is no force in that direction, that is obvious. What is not obvious at all is the next step he took, d/dt(r^2ω)= 0, and from there he goes to equating the initial angular velocity and radius with the final angular velocity and radius. Everything else from the solution falls into place from there, but as far as I see, there is a PDE, which I have no idea how to solve, and I have no idea how he implied that time derivative from that information.

Any help would be greatly appreciated and thank you in advance, I plan to repay the 'karma' when I know how to solve some physics problems with confidence on this site.

I got the problem wrong, and I am trying to understand what my teacher did. Particularly, on the right side of his solution, he took the θ acceleration to be zero, because there is no force in that direction, that is obvious.

I don't think that is obvious. A figure skater spins with increasing angular speed as he/she brings their arms in even though there is no angular force.

What is not obvious at all is the next step he took, d/dt(r^2ω)= 0, and from there he goes to equating the initial angular velocity and radius with the final angular velocity and radius.

d/dt(r^2ω)= 0 because this is a purely central force so there is no external torque.

We haven't gotten to angular momentum or central force, thanks by the way. So is what he is doing is pulling the mass out as a constant and asserting angular momentum is conserved, and if that is the case, the final angular momentum (although it's not really angular momentum because he's just using r^2w) is equal to the initial angular momentum. Man I still don't get that step, is there any way to explain it without torques and central forces and angular momentum, because we haven't even covered that yet?

If you want more help you should but a little more effort into your images. Officially we frown on posting images, but in this case I don't see a good alternative. Please improve the images, I mean, rotate the dang thing at a minimum.

But if I were given a different PDE, how would I recognize that, what is the mathematics? Also how does that information translate to the next line, where ro^2w=r^2w. Everything else in the problem just falls into place, but it seems like there is a large part of information missing in that step.

Also how does that information translate to the next line, where ro^2w=r^2w.

[itex]d/dt \left(r^2 \omega \right) = 0[/itex] means that the value of [itex]r^2 \omega[/itex] stays constant. Consequently, its initial value, [itex]r_0^2 \omega_0[/itex], is equal to the value of [itex]r^2 \omega[/itex] at any other time.

Right, I do understand that part now, mainly now it is a question of the mathematical technique used to figure out how i extrapolated d/dt(r^2w) from that partial differential equation in the first place. If I see something else, I cannot simply guess and figure out what derivative of a function is equal to the equation, you know?

Right, but if I were to come across a similar equation, is there a technique to use to figure out what the derivative of the differential equation is without guessing?

Thank you, I don't know if this will help me in subsequent work, it's not something i'd recognize, it's like a guess and check type thing, it seems there is a method of a solution, is there one, or should I just guess if i come across another problem like this on the exam?

Basically right now, I understand this problem due to you multiplying an r in there, but all I have is a memory of one problem, I don't like that, I still do not understand the method which leads to the derivative of the function being equal to the function itself. In ODEs I know you can use an integrating factor, but this is a PDE, which I have no knowledge of.

Well thanks a lot, that did increase my understanding of that problem. It is still shaky to be why I needed the derivative of the function to equal the function (and also zero because the function did), but it illuminated that the initial conditions don't change with time, so I can solve that problem, I don't know about the next one. Thank you very much though everybody.

This is, however, a "turn the crank" solution to this differential equation. "Multipltying" by [itex]dt[/itex] gives [itex]0 = r d \omega + 2 \omega d r[/itex]. Have you taken a course in ordinary differential equations? If you have, then make this equation exact, i.e., find an integrating factor.